Answer
The sequence converges to $ln2$.
Work Step by Step
Given: $a_n=ln({2n^{2}+1})-ln({n^{2}+1})$
Use logarithm quotient rule, we have
$a_n=ln(\frac{2n^{2}+1}{n^{2}+1})$
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}ln(\frac{2n^{2}+1}{n^{2}+1})$
Divide numeartor and denominator by $n^{2}$
$=\lim\limits_{n \to \infty}ln(\frac{2+1/n^{2}}{1+1/n^{2}})$
$=ln(\frac{2+\frac{1}{\infty}}{1+\frac{1}{\infty}})$
$=ln(\frac{2+0}{1+0})$
$=ln2$
Hence, the sequence converges to $ln2$.
